菜鸟的MySQL学习之旅（六）— 实战二

1. 行程和用户

Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型，枚举成员为 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。

Id	Client_Id	Driver_Id	City_Id	Status	Request_at
1	1	10	1	completed	2013-10-01
2	2	11	1	cancelled_by_driver	2013-10-01
3	3	12	6	completed	2013-10-01
4	4	13	6	cancelled_by_client	2013-10-01
5	1	10	1	completed	2013-10-02
6	2	11	6	completed	2013-10-02
7	3	12	6	completed	2013-10-02
8	2	12	12	completed	2013-10-03
9	3	10	12	completed	2013-10-03
10	4	13	12	cancelled_by_driver	2013-10-03

Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止，Role 则是一个表示（‘client’, ‘driver’, ‘partner’）的枚举类型。

Users_Id	Banned	Role
1	No	client
2	Yes	client
3	No	client
4	No	client
10	No	driver
11	No	driver
12	No	driver
13	No	driver

写一段 SQL 语句查出 2013年10月1日 至 2013年10月3日 期间非禁止用户的取消率。基于上表，你的 SQL 语句应返回如下结果，取消率（Cancellation Rate）保留两位小数。

Day	Cancellation Rate
2013-10-01	0.33
2013-10-02	0.00
2013-10-03	0.50

SELECT t.Request_at AS Day,
    ROUND(SUM(CASE WHEN t.Status= 'completed' THEN 0 ELSE 1 END)/COUNT(*),2) 
    AS 'Cancellation Rate'
FROM trips AS t
     INNER JOIN 
     users AS u ON t.Client_Id = u.Users_Id
                AND u.Banned = 'No'
GROUP BY t.Request_at;

结果

2. 各部门前3高工资的员工

将昨天employee表清空，重新插入以下数据（其实是多插入5,6两行）：

Id	Name	Salary	DepartmentId
1	Joe	70000	1
2	Henry	80000	2
3	Sam	60000	2
4	Max	90000	1
5	Janet	69000	1
6	Randy	85000	1

编写一个 SQL 查询，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：

Department	Employee	Salary
IT	Max	90000
IT	Randy	85000
IT	Joe	70000
Sales	Henry	80000
Sales	Sam	60000

• 思路
  1、链接两表，将部门id换成name
  2、如何取前三高？不妨再加入一张Employee表，将其与1中的表作对比，令e2表中的salary大于e1表中的salary
  3、限制条件：e2表中salary大于e1表中salary的个数少于3 ==》取前三高的salary

SELECT Department.Name AS Department, e1.Name AS Employee, e1.Salary AS Salary
FROM Employee e1
INNER JOIN Department
ON e1.DepartmentId = Department.Id
WHERE 3 >   (
            SELECT COUNT(DISTINCT e2.Salary) 
            FROM Employee e2
            WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId
            )
ORDER BY Department.Name, e1.Salary DESC

结果

3. 分数排名

依然是昨天的分数表，实现排名功能，但是排名是非连续的，如下：

Score	Rank
4.00	1
4.00	1
3.85	3
3.65	4
3.65	4
3.50	6

SELECT score,(
     SELECT COUNT(score) + 1 
     FROM score 
     WHERE score > s.score
) AS 'Rank'
FROM score AS s
ORDER BY Score DESC;

结果